Problem: $\dfrac{d}{dx}[\sqrt{8x^2+2x-3}]=\,?$ Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{8x+1}{\sqrt{8x^2+2x-3}}$ (Choice B) B $\dfrac{8x+1}{\sqrt{x}}$ (Choice C) C $\dfrac{1}{2\sqrt{16x+2}}$ (Choice D) D $\sqrt{16x+2}$
Solution: Since $\sqrt{8x^2+2x-3}$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. $\underbrace{\sqrt{(~\overbrace{8x^2+2x-3}^{\text{inner}}~)}}_{\text{outer}}$ So if $\sqrt{8x^2+2x-3}=w(u(x))$, then: $\begin{aligned} {u(x)}&={8x^2+2x-3} &&\text{inner function} \\\\ w(x)&=\sqrt{x}&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={16x+2} \\\\ {w'(x)}&={\dfrac{1}{2\sqrt{x}}} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} \dfrac{d}{dx}[\sqrt{8x^2+2x-3}]&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={\dfrac{1}{2\sqrt{({8x^2+2x-3})}}} \cdot {16x+2} \\\\ &=\dfrac{8x+1}{\sqrt{8x^2+2x-3}} \end{aligned}$